2013-01-03 12:47:17 UTC
I am reading "Sweet Tooth" by Ian McEwan and in the book he has a small section on the Monty Hall Problem.
Mathematicians will know this (and they are whom from whom I am seeking help [!]) but briefly; contestant chooses one of three doors, only one of which hides a prize. After his choice, Monty shows him a door with nothing behind it. The question is, is the contestant better off by switching doors?
The book has one of the most concise explanations of why he is better to switch that I have seen but then goes on to detail a story written by one of the characters. In the story a husband follows his wife to a hotel and knows that she is in one of three rooms (401, 402, 403). He picks a room in which to confront her,401, but before he can open the door, a couple come out of 402.
He knows the Monty Hall problem and switches room to 403 and finds his wife.
In the book, McEwan says that this is wrong - it is not really a Monty Hall problem as Monty *knows* which door has no prize while the couple are in their room at random (i.e. it was an accident which room they got). In order to correct the story, there should be a chambermaid who says, "I will clean the empty room and not disturb the couple"; she then enters a room and /then/ he switches.
Is McEwan right?
I don't think so; I think that Monty has no choice over which door to open (he /must/ open the unchosen door with nothing behind it; there is only one of those.)
Similarly, the couple /must/ come out of the unchosen room that doesn't contain the wife and there is only one of those. The story stands and doesn't need the chambermaid.
Is there a mathematicrat who can confirm or refute my thinking?
This is a book group book and I am going to raise this question so I'd like to have the right answer - on the other hand, if I don't find your reasoning convincing I may stick to my original thoughts.
I'm off to get a life,